Let's solve each part step by step. --- ### (a) \(\{ w \in \{a, b\}^* : w \text{ starts with } bb \text{ and } n_b(w) \equiv 1 \pmod{3} \}\) #### **Step 1: Understanding Requirements** - **Starts with "bb"**: The string must begin with two consecutive 'b's. - **\(n_b(w) \equiv 1 \pmod{3}\)**: The total number of 'b's in the string, mod 3, must be 1. #### **Step 2: Construct DFA States** Let’s break down the requirements: ##### **(i) Starts with "bb":** - States needed to ensure the string starts with "bb": - \(q_0\): Start state (no input yet). - \(q_1\): First 'b' seen. - \(q_{trap}\): Trap state (if an 'a' is seen before "bb"). - After confirming "bb" at the start, move to states that track the count of 'b's mod 3. ##### **(ii) Track \(n_b(w) \pmod{3}\):** - After "bb", we need 3 states to track the count of 'b's mod 3: - \(q_0'\): \(n_b \equiv 0 \pmod{3}\) - \(q_1'\): \(n_b \equiv 1 \pmod{3}\) (**Accept state**) - \(q_2'\): \(n_b \equiv 2 \pmod{3}\) #### **Step 3: Full DFA Construction** **States:** - \(q_0\): Start state - \(q_1\): Got 'b' - \(q_{trap}\): Trap (invalid strings) - \(q_0'\): Saw "bb", total number of b's \(\equiv 2 \pmod{3}\) - \(q_1'\): Saw "bb", total number of b's \(\equiv 0 \pmod{3}\) (since after "bb" we have 2 b's, so mod 3 = 2) - \(q_2'\): Saw "bb", total number of b's \(\equiv 1 \pmod{3}\) (**Accept state**) **Transitions:** - \(q_0 \xrightarrow{b} q_1\) - \(q_0 \xrightarrow{a} q_{trap}\) - \(q_1 \xrightarrow{b} q_0'\) (second 'b' seen, so 2 b's: \(2 \pmod{3}\)) - \(q_1 \xrightarrow{a} q_{trap}\) - \(q_{trap} \xrightarrow{a,b} q_{trap}\) - Now, from \(q_0'\) (i.e., 2 b's so far): - \(q_0' \xrightarrow{a} q_0'\) (no change to count of b's) - \(q_0' \xrightarrow{b} q_1'\) (now 3 b's: \(0 \pmod{3}\)) - \(q_1' \xrightarrow{a} q_1'\) - \(q_1' \xrightarrow{b} q_2'\) - \(q_2' \xrightarrow{a} q_2'\) - \(q_2' \xrightarrow{b} q_0'\) **Accepting State:** \(q_2'\) (**Because:** After "bb", we are at \(q_0'\) (\(2 \pmod{3}\)), and for \(n_b(w) \equiv 1 \pmod{3}\), we want the total number of b's mod 3 to be 1 at the end. As "bb" gives 2 b's, we need to accept if the final number of b's (including the first two) is 1 mod 3. That happens at \(q_2'\).) #### **Summary Table** | State | Input 'a' | Input 'b' | Description | |-----------|:---------:|:---------:|------------------------------| | q0 | trap | q1 | Start | | q1 | trap | q0' | First 'b' | | trap | trap | trap | Trap state | | q0' (2) | q0' | q1' | After "bb", 2 b's (\(2\)) | | q1' (0) | q1' | q2' | \(0 \pmod{3}\) | | q2' (1) | q2' | q0' | \(1 \pmod{3}\) (**Accept**) | **Start:** q0 **Accept:** q2' --- ### (b) \(\{b^n a^k : nk > 3\}\) #### **Step 1: Understanding Requirements** - Strings of the form \(b^n a^k\) (all b's, then all a's) - The product \(n \cdot k > 3\) #### **Step 2: DFA Construction** **Observation:** - DFA cannot count unbounded values, but \(nk > 3\) can be handled by finite states for small values. Let’s enumerate all possible values where \(n \cdot k \leq 3\) (these should **not** be accepted): - \(n = 0\) or \(k = 0\): \(nk = 0\) - \(n = 1, k = 1,2,3\): \(1,2,3\) - \(n = 2, k = 1\): \(2\) - \(n = 2, k = 2\): \(4\) (**Accept**) - \(n = 3, k = 1\): \(3\) - \(n = 4, k = 1\): \(4\) (**Accept**) So, all strings with \(n \cdot k \leq 3\) are: - \(b^n a^k\) where \(n=0\) or \(k=0\) (i.e. empty string, only b's, or only a's) - \(b a, b a^2, b a^3\) - \(b^2 a\) - \(b^3 a\) For all others, \(nk > 3\). **DFA states:** We need to distinguish \(b^n a^k\) for small \(n, k\), then send to an accept state for all larger cases. - \(q_0\): Start (no input yet) - \(q_1\): 1st 'b' - \(q_2\): 2nd 'b' - \(q_3\): 3rd 'b' - \(q_4\): 4th or more 'b's - \(q_{a0}\): 1st 'a' after n=1 - \(q_{a1}\): 2nd 'a' after n=1 - \(q_{a2}\): 3rd 'a' after n=1 - \(q_{b2a1}\): 1st 'a' after n=2 - \(q_{b2a2}\): 2nd 'a' after n=2 - \(q_{b3a1}\): 1st 'a' after n=3 - \(q_{acc}\): Accept (nk > 3) - \(q_{trap}\): Invalid (e.g., ba after a's started) **Transitions:** - \(q_0 \xrightarrow{b} q_1\), \(q_0 \xrightarrow{a} q_{trap}\) - \(q_1 \xrightarrow{b} q_2\), \(q_1 \xrightarrow{a} q_{a0}\) - \(q_2 \xrightarrow{b} q_3\), \(q_2 \xrightarrow{a} q_{b2a1}\) - \(q_3 \xrightarrow{b} q_4\), \(q_3 \xrightarrow{a} q_{b3a1}\) - \(q_4 \xrightarrow{b} q_4\), \(q_4 \xrightarrow{a} q_{acc}\) Now, for the 'a' branches: - \(q_{a0} \xrightarrow{a} q_{a1}\), \(q_{a0} \xrightarrow{b} q_{trap}\) - \(q_{a1} \xrightarrow{a} q_{a2}\), \(q_{a1} \xrightarrow{b} q_{trap}\) - \(q_{a2} \xrightarrow{a} q_{acc}\), \(q_{a2} \xrightarrow{b} q_{trap}\) - \(q_{b2a1} \xrightarrow{a} q_{b2a2}\), \(q_{b2a1} \xrightarrow{b} q_{trap}\) - \(q_{b2a2} \xrightarrow{a} q_{acc}\), \(q_{b2a2} \xrightarrow{b} q_{trap}\) - \(q_{b3a1} \xrightarrow{a} q_{acc}\), \(q_{b3a1} \xrightarrow{b} q_{trap}\) - \(q_{acc} \xrightarrow{a} q_{acc}\), \(q_{acc} \xrightarrow{b} q_{trap}\) - \(q_{trap} \xrightarrow{a,b} q_{trap}\) **Accepting State:** \(q_{acc}\) --- ## **Final Answers (Summary)** ### **(a) DFA for \(\{ w \in \{a, b\}^* : w \text{ starts with } bb \text{ and } n_b(w) \equiv 1 \pmod{3} \}\):** - **States:** \(q_0\), \(q_1\), \(q_{trap}\), \(q_0'\), \(q_1'\), \(q_2'\) - **Start:** \(q_0\) - **Accept:** \(q_2'\) - **Transitions:** As described above. ### **(b) DFA for \(\{b^n a^k : nk > 3\}\):** - **States:** \(q_0\), \(q_1\), \(q_2\), \(q_3\), \(q_4\), \(q_{a0}\), \(q_{a1}\), \(q_{a2}\), \(q_{b2a1}\), \(q_{b2a2}\), \(q_{b3a1}\), \(q_{acc}\), \(q_{trap}\) - **Start:** \(q_0\) - **Accept:** \(q_{acc}\) - **Transitions:** As described above. If you need the state diagrams, let me know!